I was working through Linear Algebra Done Right and had a question about whether a proof I gave for a question is valid or not. I've seen solutions to this question online but I wanted to see if this method worked.
Prove that if $(v_1,...,v_n)$ spans $V$, then so does the list $(v_1-v_2,v_2-v_3,...,v_{n-1}-v_n,v_n)$.
My attempt:
Let $L$ denote $(v_1-v_2,v_2-v_3,...,v_{n-1}-v_n,v_n)$ and $L_k$ be the $k$'th element of $L$.
Note that: $$v_k = \sum_{i=k}^{n} L_i$$
Let $U$ denote the vector space spanned by $(v_1-v_2,v_2-v_3,...,v_{n-1}-v_n,v_n)$.
From the sum above we can see that $v_1,...v_n\in U$. Since $U$ is a vector space every linear combination of its elements is in $U$. It follows that $V\subseteq U$, and thus $L$ spans $V$.
Does this method work or are there any tips?